taiwan-gsat 2023 Q6
8 marks
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Let $a , b , c , d , r , s , t$ all be real numbers. It is known that three non-zero vectors $\vec { u } = ( a , b , 0 )$, $\vec { v } = ( c , d , 0 )$, and $\vec { w } = ( r , s , t )$ in coordinate space satisfy the dot products $\vec { w } \cdot \vec { u } = \vec { w } \cdot \vec { v } = 0$. Consider the $3 \times 3$ matrix $A = \left[ \begin{array} { l l l } a & b & 0 \\ c & d & 0 \\ r & s & t \end{array} \right]$. Select the correct options.
(1) If $\vec { u } \cdot \vec { v } = 0$, then the determinant $\left| \begin{array} { l l } a & b \\ c & d \end{array} \right| \neq 0$
(2) If $t \neq 0$, then the determinant $\left| \begin{array} { l l } a & b \\ c & d \end{array} \right| \neq 0$
(3) If there exists a vector $\overrightarrow { w ^ { \prime } }$ satisfying $\overrightarrow { w ^ { \prime } } \cdot \vec { u } = \overrightarrow { w ^ { \prime } } \cdot \vec { v } = 0$ and cross product $\overrightarrow { w ^ { \prime } } \times \vec { w } \neq \overrightarrow { 0 }$, then the determinant $\left| \begin{array} { l l } a & b \\ c & d \end{array} \right| \neq 0$
(4) If for any three real numbers $e , f , g$, the vector $( e , f , g )$ can be expressed as a linear combination of $\vec { u } , \vec { v } , \vec { w }$, then the determinant $\left| \begin{array} { l l } a & b \\ c & d \end{array} \right| \neq 0$
(5) If the determinant $\left| \begin{array} { l l } a & b \\ c & d \end{array} \right| \neq 0$, then the determinant of $A$ is not equal to 0